A Complete Physics Resource for preparing IIT-JEE,NEET,CBSE,ICSE and IGCSE. All Physics topics are divided into multiple sub topics and are explained in detail using concept videos and synopsis.Lots of problems in each topic are solved to understand the concepts clearly.

Sunday, July 31, 2016

Energy Stored in Capacitor and Effect of Dielectric on it

Capacitor is a device used to store electric energy in it using the two plates. In storing the charge between the plates, we shall do some work and all that work done is stored in the form of energy between the plates.

To store small amount of charge, we shall do some small amount of work and it can be measured using the definition of potential. Any way that gives only small amount of work done. We need to store lot of small charges together in between the plates to get the total large charge. For doing that, we shall do large amount of work done. It is nothing but the sum of all small works done together. This can be obtained by integrating the basic equation as shown below.

Thus as shown in the above case, we can express the energy stored in a capacitor in the form of charge and potential. We can also express it in the form of capacity and voltage as shown in the above diagram.

Effect of dielectric materiel on the energy stored of a capacitor

The effect of dielectric has to be studied in two possible different cases. The first case is when the dielectric is placed and battery connected to the system is disconnected. In this case, the charge in the capacitor remains constant and we shall use the energy equation in terms of charge and capacity. Of course the charge remains same and capacity increases by dielectric constant times. As a result, the total energy of the capacitor decreases by dielectric constant times.

In the second case, the dielectric material is placed and the battery is kept connected. As battery is still connected in the system, the voltage applied to the capacitor remains constant and its capacity increases by dielectric times. Thus here we shall use the energy stored in the capacitor in terms of capacity and voltage and hence total energy stored in the capacitor increases by dielectric constant times as shown below.

Problem and Solution

Here in the given problem number of capacitors are connected in parallel and their effective capacity is given. So that we can measure the individual capacity of the capacitor. Again it is given that the capacitors are connected in series and certain voltage is applied to the system. We need to measure the total energy stored in the system.

This can be done quite easily by first measuring the effective capacity of the system when they are connected in series. Then by applying the concept of energy stored in the capacitor, we can measure the energy as shown below.

Problem and Solution

This problem is not about capacity but rather about basic potential itself. A charged particle is moving under certain voltage and we need to measure the velocity acquired by it.

It can be solved quire easily using the concept of conservation of energy as shown below. The work done due to potential difference is actually converted into kinetic energy.

Problem and Solution

This problem is also not about capacity rather about electric potential and intensity. A charged particle of known mass is attached to a string and an electric field is applied to it. Now there is force to the field and gravitational force acting on the particle simultaneously. Thus the charged particle turns an angle wit the vertical. We need to find the tension developed in the string and it can be done quite easily using the triangle law of vectors as shown in the diagram below.